化工原理第五章 精馏 答案讲课稿.doc

《化工原理第五章 精馏 答案讲课稿.doc》由会员上传分享，免费在线阅读，更多相关内容在教育资源-天天文库。

1、化工原理第五章精馏答案__________________________________________________五蒸馏习题解答1解:(1)作x-y图及t-x(y)图,作图依据如下:∵xA=(p-pB0)/(pA0-pB0);yA=pA0×xA/p以t=90℃为例,xA=(760-208.4)/(1008-208.4)=0.6898yA=1008×0.6898/760=0.9150计算结果汇总:t℃80.0290100110120130131.8x10.68980.44830.26720.12870.01950y10.91500.78750.61180.37770.072404

2、.612x/(1+3.612x)10.91120.78940.62710.40520.08400(2)用相对挥发度计算x-y值:y=αx/[1+(α-1)x]式中α=αM=1/2(α1+α2)∵α=pA0/pB0α1=760/144.8=5.249;α2=3020/760=3.974____________________________________________________________________________________________________∴αM=1/2(α1+α2)=1/2(5.249+3.974)=4.612y=4.612x/(1+3.612

3、x)由此计算x-y值亦列于计算表中,y-x图,t-x(y)图如下:1题附图2解:(1)求泡点:在泡点下两组分的蒸汽分压之和等于总压P,即:pA+pB=pA0xA+xB0xB=p求泡点要用试差法,先设泡点为87℃lgpA0=6.89740-1206.350/(87+220.237)=2.971pA0=102.971=935.41[mmHg]lgpB0=6.95334-1343.943/(87+219.337)=2.566pB0=102.566=368.13[mmHg]935.41×0.4+368.13×0.6=595≈600mmHg____________________________

4、________________________________________________________________________∴泡点为87℃,气相平衡组成为y=pA/p=pA0xA/P=935.41×0.4/600=0.624(2)求露点:露点时,液滴中参与甲苯组成应符合下列关系:xA+xB=1或pA/pA0+pB/pB0=1式中pA=0.4×760=304[mmHg];pB=0.6×760=456[mmHg]求露点亦要用试差法,先设露点为103℃,则:lgpA0=6.8974-120.635/(103+220.237)=3.165∴pA0=1462.2[mmHg]l

5、gpB0=6.95334-1343.943/(103+219.337)=2.784∴pB0=608.14[mmHg]于是:304/1462.2+456/608.14=0.96<1再设露点为102℃,同时求得pA0=1380.4;pB0=588.84304/1380.4+456/588.84=0.995≈1故露点为102℃,平衡液相组成为xA=pA/pA0=304/1380.4=0.223解:___________________________________________________________________________________________________

6、_(1)xA=(p总-pB0)/(pA0-pB0)0.4=(p总-40)/(106.7-40)∴p总=66.7KPayA=xA·pA0/p=0.4×106.7/66.7=0.64(2)α=pA0/pB0=106.7/40=2.674解:(1)yD=?αD=(y/x)A/(y/x)B=(yD/0.95)/((1-yD)/0.05)=2yD=0.974(2)L/VD=?∵V=VD+L(V/VD)=1+(L/VD)V0.96=VD0.974+L0.95(V/VD)0.96=0.974+(L/VD)0.95(1+L/VD)0.96=0.974+(L/VD)0.95(L/VD)=1.45解:简单

7、蒸馏计算:____________________________________________________________________________________________________lnW1/W2=W2=(1-1/3)W1=2/3W1;y=0.46x+0.549,x1=0.6,代入上式积分解得:釜液组成:x2=0.498,馏出液组成:WDxD=W1x1-W2x2(1/3W1)xD=W1×0.6-(2/3W1)×0.4